A horizontal force of 142 N is used to push a
47.0 kg packing crate a distance of 6.00 m on
a rough horizontal surface.
The acceleration of gravity is 9.81 m/s2 .
If the crate moves with constant velocity,
calculate
a) the work done by the force.
Answer in units of J
b) the coefficient of kinetic friction.
Why post the same question twice?
To calculate these values, we need to use the formulas for work and friction.
a) Work is defined as the product of force and displacement. When the crate moves with constant velocity, the work done by the force is equal to the work done against friction because the net work is zero. The formula to calculate work is:
Work = Force × Distance × cos(θ)
Here, θ is the angle between the force and the direction of displacement. Since the force and displacement are in the horizontal direction, θ = 0°, and cos(0°) = 1.
Therefore, the work done by the force can be calculated as:
Work = Force × Distance
Plugging in the values:
Work = 142 N × 6.00 m = 852 J
So, the work done by the force is 852 Joules (J).
b) The coefficient of kinetic friction (μk) can be determined using the following formula:
Frictional Force = μk × Normal Force
Since the crate is on a horizontal surface and not accelerating, the frictional force is equal to the force applied (142 N). The normal force is the force exerted on the crate due to gravity, which is equal to the weight of the crate. The formula to calculate weight is:
Weight = mass × acceleration due to gravity
Weight = 47.0 kg × 9.81 m/s^2 = 461.67 N
Therefore, the equation becomes:
142 N = μk × 461.67 N
Solving for μk:
μk = 142 N / 461.67 N ≈ 0.307
Hence, the coefficient of kinetic friction is approximately 0.307.